Current distribution and reduction factor for underground cables

Similarly, as fault currents are distributed through shield wires, underground cables with metallic sheath or shield earthed at both ends can also act as current carrying conductors.

The reduction factor depends on the type of cable and is given by the cable manufacturer (IEC 60909-3 2009). According to the standard SFS 6001, the chain impedance of the cable sheath and neighboring earthing grids can be considered in addition to the reduction factor, but only if the cable is significantly longer than the sections forming the chain impedance.

If the substation has an outgoing cable, according to SFS 6001 the reduction factor of the cables is to be used in place of the shield wire reduction factor. IEC 60909-3 goes deeper into how reduction factors are calculated but as they are given in the standard or from the manufacturer, this calculation is not included in this thesis. For three single-core cables, the reduction factor can even be much lower than indicated, as all 3 sheaths can contribute to distribution of fault currents (Popovic 2003). The calculation of reduction factor r3 for three single-core cables is given in the formula (4.12). However, the reduction factor for one cable sheath is often enough, as the reduction factors for cables is even smaller than for overhead line shield wires.

In calculations, a distinction between a system with three single-core cables and a three-core cable connection must be made. As the outgoing cables from a substation are typically very large in diameter, the three single-core cable situation is the usual case and is considered below. The case of three-core cables is thoroughly presented in IEC 60909-3, but not pre-sented here due to relevancy and limitations. A distinction must also be made between two fault events, a line-to-earth short-circuit in station B with the short-circuit current fed from the station A, or a short-circuit between the stations A and B. These fault events with their respective fault current distributions are explained in the next subchapters.

4.4.1 Line-to-earth short circuit in station B

Figure 4.5 illustrates how the fault currents flow in the system in a case where the fault at station B is fed from station A.

Figure 4.5 Line-to-earth short circuit in station B with fault current fed from station A only.

(IEC 60909-3 2009)

In this fault case, the current through earth that flows back to station A can be calculated with the formula as also indicated in the figure

𝐼_𝐸ẟ𝐴 = 𝑟₃3𝐼_(0)𝐴 (4.11)

where r3 is the reduction factor for the three cables, that can be calculated by the formula (4.12) given in standard IEC 60909-3, where the layout and distances between cable phases are considered, or by using the reduction factors given in the standard SFS 6001. 3I(0)A can be assumed to be equal to I”k1 in this case.

𝑟₃ = 𝑅^′_𝑆

𝑅^′_𝑆+ 3 ∙ 𝜔𝜇₀

8 + 𝑗3 ∙ 𝜔 𝜇₀ 2𝜋 ln

ẟ

√𝑟_𝑆𝑑_𝐿1𝐿2𝑑_𝐿1𝐿3

3

(4.12)

The IEC 60909-3 standard gives detailed examples for calculating the reduction factors among other things. Therefore, the exact calculations and all formulas for variables are not presented.

4.4.2 Line-to-earth short circuit between stations A and B

Figure 4.6 illustrates how the fault currents flow in the system in a case where the fault occurs between two stations and is fed from both directions.

Figure 4.6 Line-to-earth short-circuit between stations with fault current fed from stations A and B, where

I”k1 = 3I(0)A + 3I(0)B

3I(0)A = ISA + 3IEẟA

3I(0)B = ISB + 3IEẟB

(IEC 60909-3 2009)

Calculations of the current flowing through earth are considerably more complicated than in the simpler fault case presented in chapter 4.4.1. Two cases must be considered regarding earth potential rise, the current to earth at the short circuit location, and the currents through earth that flow to stations A and B.

The current to earth at the short circuit location is calculated as follows 𝐼_𝐸𝐹 = 𝑟₃3𝐼_(0)𝐴𝑍_{𝐸𝑆𝑡𝑜𝑡}

𝑅_𝐸𝐹 + 𝑟₃3𝐼_(0)𝐵𝑍_{𝐸𝑆𝑡𝑜𝑡}

𝑅_𝐸𝐹 (4.13)

where REF is the resistance at the fault point against reference earth as shown in figure 4.6 and ZEStot is calculated as follows

𝑍_{𝐸𝑆𝑡𝑜𝑡} = 1

1

𝑍^′_𝑆ℓ_𝐴+ 1

𝑍^′_𝑆ℓ_𝐵+ 1 𝑅_𝐸𝐹

= 𝑍^′_𝑆ℓ_𝐴ℓ_𝐵 ℓ +𝑍^′_𝑆ℓ_𝐴ℓ_𝐵

𝑅_𝐸𝐹

(4.14)

Equation is valid for cable lengths of at least ℓ = ẟ/2, where ẟ is the equivalent earth pene-tration depth calculated with the following formula

ẟ = 1,851 60909-3, which are presented in table 4.2. (IEC 60909-3 2009)

Table 4.2 Equivalent earth penetration depths for different soil types. (IEC 60909-3 2009)

The currents flowing to the stations A and B are calculated as follows 𝐼_𝐸ẟ𝐴 = 𝑟₃3𝐼_(0)𝐴𝑍_{𝐸𝑆𝑡𝑜𝑡}

Z’S is the self-impedance per unit length of one of the three sheaths or shields, which can be given as

R’S resistance per unit length of the sheath or shield (copper, aluminum, lead) R’S = 1/(κ‧qs) with qs ≈ 2πrsds where ds is the thickness of the sheath or shield rs is the medium radius of the sheath or shield

ẟ is the equivalent earth penetration depth presented above

Two further cases must then be considered regarding the fault point resistance as it is typi-cally unknown:

REF → ∞ no connection at the short-circuit location between sheath/shield and the soil REF → min. connection between the cable sheath/shield and the soil

(IEC 60909-3 2009) 4.4.2.1 REF → ∞

In this case the resistance between the cable sheath or shield and the soil is infinite. It is assumed that the thermoplastic sheath is not destroyed by the fault current or by the arc at the fault location. In this case the equations for earth current are as follows

𝐼_𝐸ẟ𝐴 = 𝑟₃3𝐼_(0)𝐴ℓ_𝐴

Highest currents through the sheath or shield will occur if the fault location is close to station A or station B, and the short-circuit current is fed from both directions. Similarly, the highest currents through earth are found if the fault occurs near either station. (IEC 60909-3 2009) This can also be seen from the earth current formulas (4.18) above, which are as follows when the distance ℓA or ℓB is 0

𝐼_{𝐸ẟ𝐴𝑚𝑎𝑥} = −𝑟₃3𝐼_(0)𝐵 , (ℓ_𝐴 = 0) (4.19a)

𝐼_{𝐸ẟ𝐵𝑚𝑎𝑥} = −𝑟₃3𝐼_(0)𝐴 , (ℓ_𝐵 = 0) (4.19b)

where

r3 is the reduction factor for three single-core cables 3I(0)A (ℓA = 0) is the short-circuit current fed from the station A only 4.4.2.2 REF → min.

According to IEC 60909-3, a value of 5 Ω can be used as a conservative estimate for the minimum resistance, as the connection to the surrounding earth is small. For determining the

currents through the earth, the formulas given above can be applied. As the determining current for earthing grid dimensioning is the highest current through the earth at the station, the prementioned formulas lead to

𝐼_{𝐸ẟ𝐴𝑚𝑎𝑥} = 𝑟₃3𝐼_(0)𝐴 , (ℓ_𝐴 = ℓ) (4.20a)

𝐼_{𝐸ẟ𝐵𝑚𝑎𝑥} = 𝑟₃3𝐼_(0)𝐵 , (ℓ_𝐴 = 0) (4.20b)

If the current is fed from the station A only and the current fed from the other side is ne-glected for the worst-case scenario, the current 3I(0)A (ℓA = 0) is equal to I”k1. The above design practices are valid for three single core cables regarding station earthing potential, and other practices must be followed when designing other systems with three-core cables.

More detailed formulas are available in the IEC 60909-3 standard, and only the most im-portant formulas considering this thesis’s subject area were presented.